An article reports that 1 in 500 people carry the defective gene that causes inherited colon cancer. In a sample of 2000 individuals, what is the approximate distribution of the number who carry this gene?

We can approximate this distribution by the Poisson distribution with μ = 4 .

(a) Use the Poisson approximation to calculate the approximate probability that between 3 and 7 (inclusive) carry the gene. (Round your answer to three decimal places.)

(b) Use the Poisson approximation to calculate the approximate probability that at least 7 carry the gene. (Round your answer to three decimal places.)

a) 0.71 = 71% probability that between 3 and 7 (inclusive) carry the gene.

b) 0.112 = 11.2% probability that at least 7 carry the gene.

Step-by-step explanation:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

In which

x is the number of sucesses

e = 2.71828 is the Euler number

is the mean in the given interval

In this problem, we have that:

(a) Use the Poisson approximation to calculate the approximate probability that between 3 and 7 (inclusive) carry the gene. (Round your answer to three decimal places.)

0.71 = 71% probability that between 3 and 7 (inclusive) carry the gene.

(b) Use the Poisson approximation to calculate the approximate probability that at least 7 carry the gene. (Round your answer to three decimal places.)

Either less than 7 carry the gene, or at least 7 do. The sum of the probabilities of these events is decimal 1. So

We want . So

In which

0.112 = 11.2% probability that at least 7 carry the gene.